Answer: (iv) $(100)^{(\log n)^{1/3}} + (\log \log n)^{2/3}$ is asymptotically smallest

Explanation:

Take logs or compare growth orders:

• (i) $2^n = e^{n \ln 2} = e^{\Theta(n)}$ → exponential
• (ii) $n^{\log n}= e^{(\log n)(\ln n)} = e^{\Theta((\log n)^2)}$
• (iii) $n^{\sqrt n}= e^{\sqrt n \cdot \ln n}= e^{\Theta(\sqrt n \log n)}$
• (iv) $(100)^{(\log n)^{1/3}} = e^{(\ln 100)(\log n)^{1/3}} = e^{\Theta((\log n)^{1/3})}$ plus $(\log\log n)^{2/3}= e^{\Theta(\log\log\log n)}$ which is doubly logarithmic.

For large $n$: $(\log n)^{1/3} \ll (\log n)^2 \ll \sqrt n\log n \ll n$.

Hence ordering from smallest to largest: (iv) $\ll$ (ii) $\ll$ (iii) $\ll$ (i).

Formally $f_4 = n^{o(1)}$ (sub-polynomial, quasi-polylog), $f_2 = n^{\Theta(\log n)}$ (super-polynomial but sub-exponential), $f_3 = 2^{\Theta(\sqrt n \log n)}$, $f_1=2^{\Theta(n)}$.

Correct choice: (iv)

Note: Options (ii) and (iii) are often swapped in ranking; but both dominate (iv) asymptotically.